Red-Black Trees
Comp 550
Red-black Tree
• Binary search tree + 1 bit per node: the
attribute color, which is either red or black.
• All other attributes of BSTs are inherited:
– key, left, right, and p.
• All empty trees (leaves) are colored black.
– We use a single sentinel, nil, for all the leaves of
red-black tree T, with color[nil] = black.
– The root’s parent is also nil[T ].
Comp 550
Red-black Properties
1.
2.
3.
4.
Every node is either red or black.
The root is black.
Every leaf (nil) is black.
If a node is red, then both its children are
black.
5. For each node, all paths from the node to
descendant leaves contain the same number
of black nodes.
Comp 550
Red-black Tree – Example
26
17
41
30
47
38
nil[T]
Comp 550
Remember: every
internal node has two
children, even though
nil leaves are not
usually shown.
50
Height of a Red-black Tree
• Height of a node:
– h(x) = number of edges in a longest path to a leaf.
• Black-height of a node x, bh(x):
– bh(x) = number of black nodes (including nil[T ])
on the path from x to leaf, not counting x.
• Black-height of a red-black tree is the black-height
of its root.
– By Property 5, black height is well defined.
Comp 550
Height of a Red-black Tree
h=4
26 bh=2
• Example:
• Height of a node:
17
h(x) = # of edges in a
longest path to a leaf.
• Black-height of a node
bh(x) = # of black nodes
on path from x to leaf,
not counting x.
h=1
bh=1
h=2 30
bh=1
nil[T]
Comp 550
h=1
bh=1
38
• How are they related?
– bh(x) ≤ h(x) ≤ 2 bh(x)
h=3
41 bh=2
h=2
47 bh=1
h=1 50
bh=1
Lemma “RB Height”
Consider a node x in an RB tree: The longest
descending path from x to a leaf has length h(x),
which is at most twice the length of the shortest
descending path from x to a leaf.
Proof:
# black nodes on any path from x = bh(x) (prop 5)
 # nodes on shortest path from x, s(x). (prop 1)
But, there are no consecutive red (prop 4),
and we end with black (prop 3), so h(x) ≤ 2 bh(x).
Thus, h(x) ≤ 2 s(x). QED
Comp 550
Bound on RB Tree Height
• Lemma: The subtree rooted at any node x has
 2bh(x)–1 internal nodes.
• Proof: By induction on height of x.
– Base Case: Height h(x) = 0  x is a leaf  bh(x) = 0.
Subtree has 20–1 = 0 nodes. 
– Induction Step: Height h(x) = h > 0 and bh(x) = b.
• Each child of x has height h - 1 and
black-height either b (child is red) or b - 1 (child is black).
• By ind. hyp., each child has  2bh(x)– 1–1 internal nodes.
• Subtree rooted at x has  2 (2bh(x) – 1 – 1)+1
= 2bh(x) – 1 internal nodes. (The +1 is for x itself.)
Comp 550
Bound on RB Tree Height
• Lemma: The subtree rooted at any node x has
 2bh(x)–1 internal nodes.
• Lemma 14.1: A red-black tree with n internal
nodes has height at most 2 lg(n+1).
• Proof:
– By the above lemma, n  2bh – 1,
– and since bh h/2, we have n  2h/2 – 1.
–  h  2 lg(n + 1).
Comp 550
Points to Remember
•
•
•
•
Every internal node has 2 children
But leaf nodes are typically not shown
Leaves carry no data
Red-black trees are a variation of binary
search trees to ensure that the tree is
balanced.
– Height is O(lg n), where n is the number of nodes.
• Operations take O(lg n) time in the worst case.
Comp 550
Operations on RB Trees
• All operations can be performed in O(lg n) time.
• The query operations, which don’t modify the
tree, are performed in exactly the same way as
they are in BSTs.
• Insertion and Deletion are not straightforward.
Why?
Comp 550
Rotations
Left-Rotate(T, x)
x

Right-Rotate(T, y)
y

y


Comp 550

x

Rotations
• Rotations are the basic tree-restructuring operation for
almost all balanced search trees.
• Rotation takes a red-black-tree and a node,
• Changes pointers to change the local structure, and
• Won’t violate the binary-search-tree property.
• Left rotation and right rotation are inverses.
Left-Rotate(T, x)
x

Right-Rotate(T, y)
y

y


Comp 550

x

Left Rotation – Pseudo-code
Left-Rotate (T, x)
1. y  right[x] // Set y.
2. right[x]  left[y] //Turn y’s left subtree into x’s right subtree.
3. if left[y]  nil[T ]
4.
then p[left[y]]  x
5. p[y]  p[x]
// Link x’s parent to y.
6. if p[x] = nil[T ]
7.
then root[T ]  y
Left-Rotate(T, x)
y
x
8.
else if x = left[p[x]]

x
Right-Rotate(T, y)

y
9.
then left[p[x]]  y
 
 
10.
else right[p[x]]  y
11. left[y]  x
// Put x on y’s left.
12. p[x]  y
Comp 550
Rotation
• The pseudo-code for Left-Rotate assumes that
– right[x]  nil[T ], and
– root’s parent is nil[T ].
• Left Rotation on x, makes x the left child of y, and the
left subtree of y into the right subtree of x.
• Pseudocode for Right-Rotate is symmetric: exchange
left and right everywhere.
• Time: O(1) for both Left-Rotate and Right-Rotate, since
a constant number of pointers are modified.
Comp 550
Reminder: Red-black Properties
1.
2.
3.
4.
Every node is either red or black.
The root is black.
Every leaf (nil) is black.
If a node is red, then both its children are
black.
5. For each node, all paths from the node to
descendant leaves contain the same number
of black nodes.
Comp 550
Insertion in RB Trees
• Insertion must preserve all red-black properties.
• Should an inserted node be colored Red? Black?
• Basic steps:
– Use Tree-Insert from BST (slightly modified) to insert
a node x into T.
• Procedure RB-Insert(x).
– Color the node x red.
– Fix the modified tree by re-coloring nodes and
performing rotation to preserve RB tree property.
• Procedure RB-Insert-Fixup.
Comp 550
Insertion
RB-Insert(T, z)
1.
y  nil[T]
2.
x  root[T]
3.
while x  nil[T]
4.
do y  x
5.
if key[z] < key[x]
6.
then x  left[x]
7.
else x  right[x]
8.
p[z]  y
9.
if y = nil[T]
10.
then root[T]  z
11.
else if key[z] < key[y]
12.
then left[y]  z
13.
else right[y]  z
RB-Insert(T, z) Contd.
14. left[z]  nil[T]
15. right[z]  nil[T]
16. color[z]  RED
17. RB-Insert-Fixup (T, z)
How does it differ from the
Tree-Insert procedure of BSTs?
Which of the RB properties
might be violated?
Fix the violations by calling
RB-Insert-Fixup.
Comp 550
Insertion – Fixup
• Problem: we may have one pair of consecutive reds
where we did the insertion.
• Solution: rotate it up the tree and away…
Three cases have to be handled…
http://gauss.ececs.uc.edu/Users/Franco/RedBlackTester/redblack.html
Comp 550
Insertion – Fixup
RB-Insert-Fixup (T, z)
1. while color[p[z]] = RED
2.
do if p[z] = left[p[p[z]]]
3.
then y  right[p[p[z]]]
4.
if color[y] = RED
5.
then color[p[z]]  BLACK // Case 1
6.
color[y]  BLACK
// Case 1
7.
color[p[p[z]]]  RED // Case 1
8.
z  p[p[z]]
// Case 1
Comp 550
Insertion – Fixup
RB-Insert-Fixup(T, z) (Contd.)
9.
else if z = right[p[z]] // color[y]  RED
10.
then z  p[z]
// Case 2
11.
LEFT-ROTATE(T, z)
// Case 2
12.
color[p[z]]  BLACK
// Case 3
13.
color[p[p[z]]]  RED
// Case 3
14.
RIGHT-ROTATE(T, p[p[z]]) // Case 3
15.
else (if p[z] = right[p[p[z]]])(same as 10-14
16.
with “right” and “left” exchanged)
17. color[root[T ]]  BLACK
Comp 550
Correctness
Loop invariant:
• At the start of each iteration of the while loop,
– z is red.
– If p[z] is the root, then p[z] is black.
– There is at most one red-black violation:
• Property 2: z is a red root, or
• Property 4: z and p[z] are both red.
Comp 550
Correctness – Contd.
• Initialization: 
• Termination: The loop terminates only if p[z] is black.
Hence, property 4 is OK.
The last line ensures property 2 always holds.
• Maintenance: We drop out when z is the root (since
then p[z] is sentinel nil[T ], which is black). When we
start the loop body, the only violation is of property
4.
– There are 6 cases, 3 of which are symmetric to the other 3.
We consider cases in which p[z] is a left child.
– Let y be z’s uncle (p[z]’s sibling).
Comp 550
Case 1 – uncle y is red
p[p[z]]
new z
C
C
p[z]
y
A
D
z

B



A

D

B
z is a right child here.
Similar steps if z is a left child.




• p[p[z]] (z’s grandparent) must be black, since z and p[z] are both red and
there are no other violations of property 4.
• Make p[z] and y black  now z and p[z] are not both red. But property 5
might now be violated.
• Make p[p[z]] red  restores property 5.
• The next iteration has p[p[z]] as the new z (i.e., z moves up 2 levels).
Comp 550
Case 2 – y is black, z is a right child
C
C
p[z]
p[z]
 y
A
 y
B
z

z
B



A


• Left rotate around p[z], p[z] and z switch roles  now z is a left
child, and both z and p[z] are red.
• Takes us immediately to case 3.
Comp 550
Case 3 – y is black, z is a left child
B
C
p[z]
z

A

A
 y
B

C




• Make p[z] black and p[p[z]] red.
• Then right rotate on p[p[z]]. Ensures property 4 is maintained.
• No longer have 2 reds in a row.
• p[z] is now black  no more iterations.
Comp 550
RB-Insert Example
• Case 1 (the child x, its parent and its uncle y are red):
– Change parent and uncle to black
– Change grandparent to red (preserves black height)
– Make grandparent new x, grandparent’s uncle new y
Comp 550
RB-Insert Example
• Case 2 (x and its parent are red, y is black, and x is an
“inside child”):
– Rotate outwards on x’s parent
– Former parent becomes new x
– Old x becomes parent
Comp 550
RB-Insert Example
• Case 3 (x and its parent are red, y is black, and x is an
“outside child”):
– Rotate “away from x” around x’s grandparent
– Color x’s parent black
– Now x’s old grandparent is x’s sibling; color it red to
maintain the right number of black nodes down each path
Comp 550
Algorithm Analysis
• O(lg n) time to get through RB-Insert up to the
call of RB-Insert-Fixup.
• Within RB-Insert-Fixup:
– Each iteration takes O(1) time.
– Each iteration but the last moves z up 2 levels.
– O(lg n) levels  O(lg n) time.
– Thus, insertion in a red-black tree takes O(lg n) time.
– Note: there are at most 2 rotations overall.
Comp 550
Deletion
• Deletion, like insertion, should preserve all the
RB properties.
• The properties that may be violated depends
on the color of the deleted node.
– Red – OK. Why?
– Black?
• Steps:
– Do regular BST deletion.
– Fix any violations of RB properties that may result.
Comp 550
Deletion
RB-Delete(T, z)
1. if left[z] = nil[T] or right[z] = nil[T]
2.
then y  z
3.
else y  TREE-SUCCESSOR(z)
4. if left[y] = nil[T ]
5.
then x  left[y]
6.
else x  right[y]
7. p[x]  p[y] // Do this, even if x is nil[T]
Comp 550
Deletion
RB-Delete (T, z) (Contd.)
8. if p[y] = nil[T ]
9. then root[T ]  x
10. else if y = left[p[y]]
11.
then left[p[y]]  x
12.
else right[p[y]]  x
13. if y = z
14. then key[z]  key[y]
15. copy y’s satellite data into z
16. if color[y] = BLACK
17. then RB-Delete-Fixup(T, x)
18. return y
Comp 550
The node passed to
the fixup routine is
the lone child of the
spliced up node, or
the sentinel.
RB Properties Violation
• If y is black, we could have violations of redblack properties:
– Prop. 1. OK.
– Prop. 2. If y is the root and x is red, then the root
has become red.
– Prop. 3. OK.
– Prop. 4. Violation if p[y] and x are both red.
– Prop. 5. Any path containing y now has 1 fewer
black node.
Comp 550
RB Properties Violation
• Prop. 5. Any path containing y now has 1 fewer black
node.
–
–
–
–
Correct by giving x an “extra black.”
Add 1 to count of black nodes on paths containing x.
Now property 5 is OK, but property 1 is not.
x is either doubly black (if color[x] = BLACK) or red & black
(if color[x] = RED).
– The attribute color[x] is still either RED or BLACK. No new
values for color attribute.
– In other words, the extra blackness on a node is by virtue
of x pointing to the node.
• Remove the violations by calling RB-Delete-Fixup.
Comp 550
Deletion – Fixup
RB-Delete-Fixup(T, x)
1. while x  root[T ] and color[x] = BLACK
2.
do if x = left[p[x]]
3.
then w  right[p[x]]
4.
if color[w] = RED
5.
then color[w]  BLACK
// Case 1
6.
color[p[x]]  RED
// Case 1
7.
LEFT-ROTATE(T, p[x])
// Case 1
8.
w  right[p[x]]
// Case 1
Comp 550
RB-Delete-Fixup(T, x) (Contd.)
/* x is still left[p[x]] */
9.
if color[left[w]] = BLACK and color[right[w]] = BLACK
10.
then color[w]  RED
// Case 2
11.
x  p[x]
// Case 2
12.
else if color[right[w]] = BLACK
13.
then color[left[w]]  BLACK // Case 3
14.
color[w]  RED
// Case 3
15.
RIGHT-ROTATE(T,w)
// Case 3
16.
w  right[p[x]]
// Case 3
17.
color[w]  color[p[x]]
// Case 4
18.
color[p[x]]  BLACK
// Case 4
19.
color[right[w]]  BLACK
// Case 4
20.
LEFT-ROTATE(T, p[x])
// Case 4
21.
x  root[T ]
// Case 4
22.
else (same as then clause with “right” and “left” exchanged)
23. color[x]  BLACK
Comp 550
Deletion – Fixup
• Idea: Move the extra black up the tree until x points to a
red & black node  turn it into a black node,
• x points to the root  just remove the extra black, or
• We can do certain rotations and recolorings and finish.
• Within the while loop:
– x always points to a nonroot doubly black node.
– w is x’s sibling.
– w cannot be nil[T ], since that would violate property 5 at p[x].
• 8 cases in all, 4 of which are symmetric to the other.
Comp 550
Case 1 – w is red
p[x]
B
w
x
A

D
B
D

C

x A
E




E

new w C




• w must have black children.
• Make w black and p[x] red (because w is red p[x] couldn’t have
been red).
• Then left rotate on p[x].
• New sibling of x was a child of w before rotation  must be
black.
• Go immediately to case 2, 3, or 4.
Comp 550
Case 2 – w is black, both w’s children are
black
p[x]
new x
B
c
w
x
A

D
A


C

•
•
•
•
B c
E


D

C


E


Take 1 black off x ( singly black) and off w ( red).
Move that black to p[x].
Do the next iteration with p[x] as the new x.
If entered this case from case 1, then p[x] was red  new x is
red & black  color attribute of new x is RED  loop
terminates. Then new x is made black in the last line.
Comp 550

Case 3 – w is black, w’s left child is red, w’s
right child is black
B
c
w
x
A

B c
x
D


C

E


A
C

new w
D



E

• Make w red and w’s left child black.
• Then right rotate on w.
• New sibling w of x is black with a red right child  case 4.
Comp 550

Case 4 – w is black, w’s right child is red
B
c
w
x
A

D
B
D

C

c’

x A
E

E


•
•
•
•

C




Make w be p[x]’s color (c).
Make p[x] black and w’s right child black.
Then left rotate on p[x].
Remove the extra black on x ( x is now singly black) without
violating any red-black properties.
• All done. Setting x to root causes the loop to terminate.
Comp 550
Explanation of RB-Delete-Fixup
• x is the current node passed to the RB-Delete-Fixup, which is called within RBDelete. Darkened nodes are black, heavily shaded nodes are red, lightly shaded
nodes (represented as c and c’) are either red or black.
• w is a sibling of x. α,β, …, ζ represent arbitrary sub-trees.
• In each case (1, 2, 3, and 4), the before configuration is transformed to the after
configuration by changing color and/or performing rotation.
• Case 1 is transformed into case 2, 3, or 4 by exchanging the colors of nodes B and
D, performing a left rotation.
• In case 2, extra black represented by pointer x is moved up the tree by coloring
node D red and pointing x to B
• Case 3 is transformed to case 4 by exchanging color of C and D, plus performing a
right rotation.
• In case 4, extra black represented by x can be moved by exchanging colors, and
performing a left rotation. Then the loop terminates.
Comp 550
Analysis
• O(lg n) time to get through RB-Delete up to
the call of RB-Delete-Fixup.
• Within RB-Delete-Fixup:
– Case 2 is the only case in which more iterations
occur.
• x moves up 1 level.
• Hence, O(lg n) iterations.
– Each of cases 1, 3, and 4 has 1 rotation   3
rotations in all.
– Hence, O(lg n) time.
Comp 550